Separable First Order Differential Equations: Solving y'=x√y

[dragonblood]

I have tried to solve the differential equation

$$y'=x\sqrt{y}$$

like this:

$$y^{-\frac{1}{2}}y'=x$$
$$\int{y^{-\frac{1}{2}}}dy=\int{xdx}$$
$$y^{\frac{1}{2}}=\frac{x^2 +C}{4}$$
$$y=\left(\frac{x^2+C}{4}\right)^2$$

Is this the right way to solve it? Because the answer in my textbook says that the answer is
$$y=\pm\sqrt{x^2+C}$$

But I really can't see where I've gone wrong.

 

[arildno]

The textbook's answer solves the diff.eq y'=x/y, rather than the one given.

See if your book may have mixed up the ordering of solutions to various diff.eq problems!

(Your own solution is correct for the problem given)

 

[CFDFEAGURU]

Yes, I got the same solution as you did, dragonblood.

Wow, never thought I could discuss ODEs with someone named dragonblood.

Cool lol

Matt

 

[arildno]

You wouldn't want to know what arildno means in Norwegian, SMLSKDMGLGURU!